Calculating Radioactive Decay: Estimating Age of Ancient City

[ussrasu]

Q: A 50-g chunk of charcoal is found in the buried remains of an ancient city destroyed by invaders. The carbon-14 activity of the sample is 200 decays/min. Roughly when was the city destroyed?

A: I used R = Ro*e^(-lambda*t) and t1/2 = ln2/lambda

t1/2 = 3.834*10^-12 s-1

and 200 decays/min = 3.33 decays/sec or 3.33 Bq

What is the initial acitivity/kg (Ro)?
The final acitivty/kg = 3.33 Bq/0.05 Kg = 66.66 Bq/Kg

Therefore R/Ro = 66.66/255

I then found the time to be 11 088 years

Can someone please check over this and see if this looks correct and advise of any corrections?

Many thanks!

 

[vanesch]

I didn't check the numbers (for instance, is the initial activity of C-14 255 Bq/kg ?) but the approach seems totally correct.

cheers,
Patrick.

 

[Astronuc]

What half-life of C-14 is used?

Be careful with units. t_1/2 should have units of time.

1 Bq = 1 decay/s is correct.

The decay constant $\lambda$ = ln 2 / t_1/2, and customarily it is often expressed as s^-1, but it should be compatible with whatever units (s, hr, days) are used for time, t.

I calculate a decay constant of 3.9455E-12 s^-1.

Activity A(t) = A_o exp(-$\lambda$t)

So inital activity is A(t) exp ($\lambda$t).

So to solve the problem, one needs to know the normal specific activity for C-14 in charcoal. Where did 255 Bq/kg originate?

255/66.66 = 3.825 half-lives.

http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/halfli2.html

http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/halfli.html

Useful calculator - http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/raddec.html#c1

 

[ussrasu]

In the textbook in a previous example - it says that the C-14 activity in a living tree is 255 Bq - that is where i got that value from but i wasnt sure if i could use that value in this question - ie is it always 255 Bq in living trees?!?

The half-life is in s^-1 - therefore the answer would be in seconds and i just converted it back to years in the answer.

Thanks